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For a fixed pole away from the saddle point, the smooth amplitude there is $(i-z_0{)}^{-1}$. The simple-saddle contribution in steepest descent isThe residue theorem supplies an additional contribution precisely when the contour deformation crosses $z_0$. The region swept above the original contour has positive orientation: the original contour from left to right followed by the reversed saddle contour encloses it counterclockwise. Consequently the original integral equals the saddle integral plus the pole contribution:For $|z_0|<1$, an upper-half-plane pole outside the indentation has $\chi =1$, while a lower-half-plane pole has $\chi =0$. The entire upper unit half-disk lies below the saddle parabola, so there is no further case inside that half-disk. With a fixed indentation radius $\delta$, a pole inside the upper semicircle, $|z_0|<\delta$, is also below the original contour and has $\chi =0$; real points in the indentation gap are excluded as well. Taking $\delta <|z_0|$ for a fixed nonzero pole gives the usual upper/lower classification. The pole contribution need not always dominate exponentially; its size depends on $\mathrm{Re}\varphi (z_0)$, so keeping both terms makes that dependence explicit.

A pole on the original contour requires a stated Cauchy principal value or an indentation prescription. A pole parameter on the negative imaginary branch cut still defines the integral: the contour stays on its fixed sheet, and only the denominator uses $z_0$. Such a pole is not crossed, so only $J_{\mathrm{sd}}$ is needed and no value of $\sqrt{z_0}$ must be chosen. The residue exponential above is evaluated only when $\chi =1$. At $z_0=0$ the original indentation excludes the singular point, so no additional residue is crossed. These conventions matter before taking any limiting pole position.

Write the phase and amplitude asThe saddle point relevant to the contour is $z=0$, since ${\varphi }^{\prime }(z)=\sinh z$ andThe local steepest-descent direction is therefore vertical: with $z=iy$,

The given contour approaches the lines $\mathrm{Im}z=-\pi$ at its left end and $\mathrm{Im}z=\pi$ at its right end. The poles of $f$ are at $z=\pm 4i$, outside the closed strip $|\mathrm{Im}z|\le \pi$. The Cauchy integral theorem therefore permits deformation to the contour consisting of the lower horizontal ray from $-\infty -i\pi$ to $-i\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the upper horizontal ray from $i\pi$ to $\infty +i\pi$. On either horizontal ray,so those two integrals are exponentially small, of order $e^{-x}{x}^{-1/2}$.

The vertical segment, oriented upward, contributesIts real phase has a unique maximum at $y=0$. Sincethe simple-saddle contribution in steepest descent, equivalently the local Gaussian integral, givesThe factor $i$ comes from the upward tangent $dz=idy$ of the deformed contour.

Write $w=z-1=X+iY$. The phase and its derivative areso the saddle points arewith $\varphi (z_{+})=2i$ and $\varphi (z_{-})=-2i$. The contour geometry can be drawn fromThe stationary-phase level $\mathrm{Re}\varphi =0$ consists of $X=0$ and $X^2-3Y^2+3=0$, meeting at the saddles. The steepest curves through $z_{\pm }$ are the levels $\mathrm{Im}\varphi =\pm 2$. Far away, sectors with $\cos (3\arg w)>0$ are exponential hills and those with $\cos (3\arg w)<0$ are valleys.

The stated contour deforms through the upper saddle. If $z=z_{+}+s$, thenThe descent tangent has $s=e^{i\pi /4}t$, because $3i{s}^2=-3t^2$. The simple-saddle contribution in steepest descent is therefore

When the contour begins at $z=1$, deform it first from the endpoint into the decaying negative-real direction and then onto the same upper-saddle descent path. Near the endpoint, $s=z-1$ and $\varphi =3s+O(s^3)$, so the endpoint contribution in steepest descent isAdding the saddle and endpoint pieces gives

Write the contour near the simple saddle aswith its given orientation. Since ${\varphi }^{\prime }(z_0)=0$,Along a steepest-descent tangent the quadratic coefficient is real and negative, soMoreover $dz=e^{i\alpha }ds$. Only a neighbourhood of width $O(x^{-1/2})$ contributes at leading order, and there $f(z)=f(z_0)+o(1)$. The local integral is consequentlyEvaluating the Gaussian proves the simple-saddle contribution in steepest descent: